Đặt Rx = x
\(Ront\left(R1\backslash\backslash Rx\right)\Rightarrow R_{tđ}=R_o+\dfrac{R_1x}{R_1+x}=4+\dfrac{12x}{12+x}=\dfrac{16x+48}{12+x}\)
\(\Rightarrow I=\dfrac{U}{R_{tđ}}=\dfrac{16}{\dfrac{16x+48}{12+x}}=\dfrac{12+x}{x+3}\)
\(\Rightarrow U_{R_x}=U_{R_1}=U_{R_{1x}}=\dfrac{12+x}{x+3}.\dfrac{12x}{12+x}=\dfrac{12x}{x+3}\)
Mà: \(\dfrac{U_{R_x}^2}{R_x}=9\)\(\Leftrightarrow\dfrac{\dfrac{144x^2}{\left(x+3\right)^2}}{x}=9\)\(\Leftrightarrow\dfrac{144x}{x^2+6x+9}=9\)
\(\Leftrightarrow9x^2-90x+81=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Hiệu xuất của mạch điện: \(H=\dfrac{R_{1x}.I^2}{R.I^2}.100\%=\dfrac{R_{1x}}{R}.100\%=\dfrac{3R_x}{4\left(3+R_x\right)}.100\%\)
Xét với Rx = 1 => H1 = 18,75%
Xét với Rx = 9 => H2 = 56,25%
b, Từ câu a
=> \(P_{R_x}=\dfrac{U^2}{R_x}=\dfrac{\left(\dfrac{12x}{x+3}\right)^2}{x}=\dfrac{144x}{x^2+6x+9}=\dfrac{144}{x+\dfrac{9}{x}+6}\)
Ta có để Prx Max => x+9/x + 6 Min
=> x + 9/x Min.
Áp dụng baddt Côsi có: \(x+\dfrac{9}{x}\ge2\sqrt{x.\dfrac{9}{x}}=6\)
=> \(MAX.P_{R_x}=\dfrac{144}{6+6}=12\left(W\right)\)
Vậy....................
Bye khò khò